---
title: "Transformer on load, phasor diagrams"
subject: "Electrical Machine"
chapter: 2
key: "semester_4/electrical_machine:transformer-phasor-resistive-capacitive"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/transformer-phasor-resistive-capacitive
site: BCT Engineering
---

# Transformer on load, phasor diagrams

Electrical Machine · Chapter 2: Transformer

- Asked in: 2068 Magh, 8 marks
- Video: [How to Draw Transformer Phasor Diagram](https://www.youtube.com/watch?v=Y4FLaqdNaTw) — Study EETech
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/transformer-phasor-resistive-capacitive
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-2

## Question

Explain the transformer on load and no load with the phasor diagram of resistive and capacitive load.

## Short answer

At no load, flux Φ lags V1 by 90°; I0 lags V1 by about 80°. With resistive load, I2′ is in phase with V2, so I1 = I0 + I2′ lags V1 slightly. With capacitive load, I2′ leads V2, so I1 can lead V1 at high load, and V2 may rise above no-load value.

## Answer

A transformer has a primary and secondary with N1 and N2 turns. Its behaviour at no load and with loads of different power factors is described below. Quantities on the secondary are referred to the primary with a dash (e.g. I2′ = K I2).

#### No-load operation
- Secondary open, so I2 = 0. The primary takes the no-load current I0 = Iμ + Iw, where Iw (core-loss component, in phase with V1) and Iμ (magnetising component, in phase with Φ, lagging V1 by 90°).
- The flux Φ lags V1 by about 90°. The emfs E1 and E2 lag Φ by 90°. E1 is nearly equal and opposite to V1.
- I0 lags V1 by an angle φ0 (typically 75° to 85°), so the no-load power factor is poor.

**Phasor construction (no-load), in order:** (1) draw Φ along the reference axis; (2) draw Iμ along Φ; (3) draw E1 and E2 lagging Φ by 90°; (4) V1 = −E1 (equal and opposite); (5) draw Iw along V1; (6) I0 = Iμ + Iw, at angle φ0 behind V1.

#### Loaded operation
The secondary current I2 flows, producing an mmf that opposes the flux. The primary draws the extra current I2′ = (N2/N1) I2 (opposite in phase to I2). The total primary current is I1 = I0 + I2′ (phasor sum). Φ stays constant.

For the ideal-load phasor diagram, neglect winding resistance and leakage, so V2 = E2.

| Step | Resistive load (pf = 1) | Capacitive load (leading pf) |
|---|---|---|
| Reference | V2 = E2 | V2 = E2 |
| I2 | In phase with V2 | Leads V2 by φ2 |
| I2′ | Equal in magnitude to K I2, opposite to I2 | Same, reversed (180°) |
| I1 = I0 + I2′ | I2′ is in phase with V1, so I1 is the sum of I2′ and I0 and lags V1 by a small angle | I2′ leads V1, so I1 leads V1 if I2′ is larger than the lagging part of I0 |
| Angle φ1 between V1 and I1 | Small, I1 nearly in phase with V1 | Leading when I2′ is large compared with I0 |
| Primary pf | Near unity (lagging) | Leading (lagging at light load) |

**Remarks.** With a resistive load the primary current is almost in phase with V1; with a capacitive load the secondary current leads V2 by φ2, so I1 also leads V1. In the actual transformer, the voltage drops I1R1, I1X1, I2R2 and I2X2 are subtracted from E1 and E2. Because of these drops V2 falls for lagging and unity loads, but may **rise** above its no-load value for a leading (capacitive) load (negative regulation).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How does flux relate to V1 at no-load?
   A. Flux leads V1 by 90 degrees
   B. Flux lags V1 by about 90 degrees
   C. Flux is 180 degrees from V1
   D. Flux is in phase with V1
   Answer: B. E1 lags flux by 90 degrees and V1 = -E1, so flux lags V1 by about 90 degrees. (Key point: No-load phasor)

2. How is the total primary current found on load?
   A. I1 = I0 + I2' as a phasor sum
   B. I1 = I2/K
   C. I1 = I2'
   D. I1 = I0 - I2
   Answer: A. Primary carries no-load current plus the load component I2' = K I2. (Key point: Load current)

3. With a resistive load, how does I1 compare with V1?
   A. Lagging by 90 degrees
   B. Opposite
   C. Nearly in phase
   D. Leading by 90 degrees
   Answer: C. I2' is in phase with V1, so I1 is near unity pf, lagging slightly due to I0. (Key point: Resistive load)

4. What can happen to V2 with a leading load?
   A. It always falls
   B. It may rise above its no-load value (negative regulation)
   C. It stays equal
   D. It is zero
   Answer: B. A leading current gives a rise in terminal voltage because of the winding drops. (Key point: Capacitive load regulation)

## Related in Electrical Machine

- [Transformer on load and no load, phasor diagrams](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atransformer-load-phasor-resistive-capacitive) — Electrical Machines, chapter 2
- [Transformer at resistive, inductive, capacitive loads](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atransformer-loading-conditions-phasor) — Electrical Machines, chapter 2
- [Loaded transformer and phasor diagram for inductive load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Atransformer-loaded-operation-phasor-inductive) — Electrical Machines I, chapter 2
- [Practical vs ideal transformer, phasor diagram](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Apractical-vs-ideal-transformer-phasor) — Electrical Machines, chapter 2
- [Transformer no-load and load operation, constant core flux](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aideal-transformer-noload-load-constant-flux) — Electrical Machines I, chapter 2, asked 5×
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## Related in other subjects

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